Extrinsic Semiconductors Extrinsic semiconductors : impurity atoms dictate the properties Almost all commercial semiconductors are extrinsic Impurity concentrations of 1 atom in 1012 is enough to make silicon extrinsic at room T! Impurity atoms can create states that are in the band gap. In most cases, the doping of a semiconductor leads either to the creation of donor or acceptor levels n-type semiconductors n-Type In these, the charge carriers are negative p-type semiconductors. In these, the charge carriers are positive p-Type Band Diagram: Acceptor Dopant in Semiconductor For Si, add a group III element to “accept” an electron and make p-type Si (more positive “holes”). EC “Missing” electron results in an extra “hole”, with an acceptor energy level E just above E A F E the valence band E . A V EV p-type Si Holes easily formed in valence band, greatly increasing the electrical conductivity. Fermi level EF moves down towards EV. Typical acceptor elements are Boron, Aluminum, Gallium, Indium. Band Diagram: Donor Dopant in Semiconductor Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) For group IV Si, add a group V element to “donate” an electron and make n-type Si n-type Si (more negative electrons!). EC ED “Extra” electron is weakly bound, with EF donor energy level ED just below conduction band EC. Egap~ 1 eV EV Dopant electrons easily promoted to conduction band, increasing electrical conductivity by increasing carrier density n. Fermi level EF moves up towards EC. Typical donor elements that are added to Si or Ge are phosphorus, arsenic, antimonium. Silicon n-type semiconductors: Si Si Si Si Bonding model description: Element with 5 bonding electrons. Only 4 electrons Si Si Si Si - participate in bonding the extra e can easily Si P Si Si become a conduction electron Si Si Si Si p-type semiconductors: Si Si Si Si Bonding model description: Element with 3 bonding electrons. Since 4 Si Si Si Si electrons participate in bonding and only 3 are Si Si B Si available the left over “hole” can carry charge Si Si Si Si The Mass Action Law This relationship is valid for both intrinsic and extrinsic semiconductors. In an extrinsic semiconductor the increase in one type of carrier (n or p) reduces the concentration of the other through recombination so that the product of the two (n and p) is a constant at a any given temperature. The carriers whose concentration in extrinsic semiconductors is the larger are designated the majority carriers, and those whose concentration is the smaller the minority carriers. At equilibrium, with no external influences such as light sources or applied voltages, the concentration of electrons,n0, and the concentration of holes, p0, are related by 2 ni denotes the carrier concentration in no× p o = n i intrinsic silicon A material is defined as intrinsic when it consists purely of one element and no outside force (like light energy) affects the number of free carrier other than heat energy. In intrinsic Si, the heat energy available at room temperature generates approximately 1.5x1010 carriers per cm3 of each type (holes and electrons) . The number of free carriers doubles for approximately every 11°C increase in temperature. This number represents a very important constant (at room temperature), and we define 10 -3 ni = 1.5x10 cm where ni denotes the carrier concentration in intrinsic silicon at room temperature (constant for a given temperature). Based on charge neutrality, for a sample doped with ND donor -3 -3 atoms per cm and NA acceptor atoms per cm we can write no + NA = po + ND which shows that the sum of the electron concentration plus the ionized acceptor atoms is equal to the sum of the hole concentration plus the ionized donor atoms. The equation assumes that all donors and acceptors are fully ionized, which is generally true at or above room temperature. Given the impurity concentration, the above equations can be solved simultaneously to determine electron and hole concentrations. In electronic devices, we typically add only one type of impurity within a given area to form either n-type or p-type regions. In n-type regions there are typically only donor impurities and the donor concentration is much greater than the intrinsic carrier concentration, NA=0 and ND>>ni. Under these conditions we can write no ≈ ND where no is the free electron concentration in the n-type material and 3 ND is the donor concentration (number of added impurity atoms/cm ). Since there are many extra electrons in n-type material due to donor impurities, the number of holes will be much less than in intrinsic silicon and is given by, 2 po = ni / ND where po is the hole concentration in an n-type material and ni is the intrinsic carrier concentration in silicon. Similarly, in p-type regions we can generally assume that ND=0 and NA>>ni. In p-type regions, the concentration of positive carriers (holes), po, will be approximately equal to the acceptor concentration, NA. po = NA and the number of negative carriers in the p-type material, no, is given by 2 no = ni / NA Notice the use of notation, where negative charged carriers are n, positive charged carriers are p, and the subscripts denote the material, either n-type or p-type. This notation will be used throughout our discussion of p-n junctions and bipolar transistors. The above relationships are only valid when ND or NA is >> ni, which will always be the case in the problems related to integrated circuit design. Example Calculate the conductivity and the resistivity of a n-type silicon wafer which contains 1016 electrons per cubic centimeter with an electron mobility of 1400 cm2/Vs. Solution: The conductivity is obtained by adding the product of the electronic charge, q, the carrier mobility, and the density of carriers of each carrier type, or: σ q= (μn n+ μ p ) p As n-type material contains almost no holes, the conductivity equals: -19 16 σ= q μn n = 1.6 x 10 x 1400 x 10 = 2.24 1/Ωcm. The resistivity quals the inverse of the conductivity or: e 1 1 ρ= = σq() μn n+ p μ p and equals ρ = 1/σ = 1/2.24 = 0.446 Ωcm. Example A Si sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of donors at room temperature, calculate the charge carrier concentration and conductivity at room temperature. -3 2 -1 -1 2 -1 -1 [For Si: ρ = 2330 kg m , atomic weight = 28, μe = 0.15 m V s , μh = 0.05 m V s , 10 3 ni= 1.5x10 carriers per cm ] Solution: 1) Calculate the fraction of donor atoms (phosphorus atoms per silicon atom) where N – number of Si atoms per unit volume N Si D =10−6 N 2) Calculate the number of silicon atoms per unit volume Si −3 ρ 2330Kg. m 23 −1 28 3 N=Si ×NAvogadro = −3 −×16 ×atoms 10 . =5 molSi × 10 atoms − . m ASi 28× Kg 10 . mol 22 −3 3) Calculate the number of donors atoms (phosphorus) N=D 5P × 10 atoms − . m 4) As NA=0 and ND>>ni , then we can safely assume that no=ND and po is very small ~ zero =σ no × q μ ×e =σ 5()P × 1022 atoms −)(1..×3 6 m × 10)(19C . × m02 -1 V15 -1 s σ=1200 − Ω1.m − 1 Consider the following equations: Mass Action: Neutrality: 2 no× p o = n i no+ N A= p o+ DN Solving the equations simultaneously Electron concentration (n-type semiconductor) Hole concentration (p-type semiconductor) 2 2 1 ⎡ 2 2 ⎤ 1 ⎡ ⎤ n=() N − N + N () − N4p + = n o () N2 −AD N +AD N () − N4 i + n o 2 ⎣⎢ DA DA i ⎦⎥ ⎣⎢ ⎦⎥ 2 2 ni ni no = po = po no Solving the problem again. NA=0 n=() N1 ⎡ − N + N () −2 N4 2 +⎤ n o 2 ⎣⎢ DA DA i ⎦⎥ 2 2 1 ⎡ 22 22 2 16 2 ⎤ n=1 N⎡ + N5=4 + ⎤ 10 n × + 5( 10 ×) ( + 4. 1) 5 × 10 o 2 ⎣⎢ D D i ⎦⎥ 2 ⎣⎢ ⎦⎥ 22 no=5 × 10 2 16 2 ni 1() 5. × 10 10− 3 po= = 220= 45. × 10m no 5× 10 =σ q ×o() n ×μ e + o p μ h × )(=1σ 6 ×. 1019C ×[] (P 5 ×22 10 atoms. ( − 3 .0× )m m 152 -1V+0 -1 s 45 () × . 10 10 () ×0.05 ) σ=1200 − Ω1.m − 1 Example An n-type piece of silicon of length L = 10 micron has a cross sectional area A = 0.001 cm2. A voltage V = 10 Volt is applied across the sample yielding a current I = 100 mA. What is the resistance, R of the silicon sample, its conductivity, σ, and electron density, n ? 2 μn= 1400 cm /Vs Solution The resistance of the sample equals R = V/I = 10/0.1 = 100 Ω. Since R = L /(σA) the conductivity is obtained from: σ = L/(R A) = 0.001/(100 x 0.001) = 0.01 1/Ωcm. The required electron density is related to the conductivity by: σ = q n μ n so that the density equals: -19 13 -3 n = σ/(q μ n) = 0.01/(1.6 x 10 x 1400) = 4.46 x 10 cm . Example A Si sample at room temperature is doped with 1011 As atoms/cm3. What are the equilibrium electron and hole concentrations at 300 K? Solution 2 Since the NA is zero we can write, no po = ni And no + NA = po + ND 2 2 → no –ND no –ni = 0 11 -3 Solving this quadratic equations results in n0 = 1.02x10 [cm ] 2 20 11 and thus, p0 = ni / n0 = 2.25x10 / 1.02x10 9 -3 p0 = 2.2x10 [cm ] Notice that, since ND>ni, the results would be very similar if we assumed 11 -3 no=ND=10 cm , although there would be a slight error since ND is not much greater than ni.